Math Question and Answers


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 66851 by John Kaloki Musau last updated on 20/Aug/19

Commented by John Kaloki Musau last updated on 20/Aug/19
A piece of wire, pcm long, is bent to form the shape shown in the figure above. The figure consists of a semicircular arc of radius rcm and two perpendicular sides of length xcm each. Express x in terms of p and r, hence show that the area of the figure is given by A=1/2πr^2+1/8(p-πr)^2
	Answered by John Kaloki Musau last updated on 20/Aug/19

	$T h e a n s w e r s a r e,$ $(i) x = \frac{p - π r}{2} c m$ $(i i) A = \frac{1}{2} π r^{2} + \frac{1}{2} \times x \times x = \frac{1}{2} π r + \frac{1}{2} {(\frac{p - π r}{3})}^{2}$ $T h o s e a r e t h e a n s w e r s b u t I$ ${d o n}^{'} t k n o w h o w t o c a l c u l a t e$ $t h e m . P l e a s e s h o w m e .$
	Answered by $@ty@m123 last updated on 20/Aug/19

	$L e n g t h o f h a l f c i r c u m f e r e n c e$ $= π r (f o r m u l a)$ $∴ L e n g t h o f w i r e = x + x + π r$ $\Rightarrow p = 2 x + π r$ $\Rightarrow 2 x = p - π r$ $\Rightarrow x = \frac{p - π r}{2}$ $- - - - - - - - - - - - - - - - -$ $A r e a o f s e m i c i r c l e = \frac{π r^{2}}{2} . . . . (1) (f o r m u l a)$ $A r e a o f t r i a n g l e = \frac{1}{2} \times b a s e \times h e i g h t (f o r m u l a)$ $A r e a o f t r i a n g l e = \frac{1}{2} \times {(\frac{p - π r}{2})}^{2} . . . . (2)$ $∴ A r e a o f s h a p e = (1) + (2)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum